According to Biot-Savart’s law, magnetic field due to a straight current carrying wire at a point, at a distance r from it, is given by `B=(mu_(0)I)/(4pir)( sin theta_(1)+sin theta_(2))`. The direction of magnetic field being perpendicular to the plane containing the wire and that point.

In the figure shown, a closed loop AOCBA in which current I is flowing as shown. Given OA = OB = OC = a. Find the magnetic field at point B due to the loop.

Options

A. `-(mu_(0)I)/(4pia)(hati+hatk)`
B. `-(mu_(0)I)/(4pia)(hatj+hatk)`
C. `(-mu_(0)I)/(2pia)(hatj+hatk)`
D. `(-mu_(0)I)/(2sqrt(2)pia)(hatj+hatk)`

Answer

C

Explanation

At O : Magnetic field due to AO and OC will be zero.
Due to `CB:r=a cos 45^(@)=a/(sqrt(2))`
`vecB_(CB)=(mu_(0))/(4pi)I/r[sin 45^(@)+sin45^(@)](-hatk)=(-mu_(0)I)/(2pia)hatk`
similarly, `vecB_(BA)=(-mu_(0))/(2pi)I/a hatj :. vecB_("net")=vecB_(CB)+vecB_(BA)=(-mu_(0))/(2pi) I/a(hatj+hatk)`